The system can be written as

A X = 0

Now, |A| = 3(12 – 21) + 1(16 – 15) + 2(28 – 15)

|A| = – 27 + 1 + 26

|A| = 0

Hence, the system has infinite solutions

Let z = k

3x – y = – 2k

4x + 3y = – 3k

A X = B

|A| = 9 + 4 = 13 ≠0 So, A ^{– 1} exist

Now adj A = =

X = A ^{– 1} B =

X =

Hence, x = , y = and z = k